3.2.16 \(\int \frac {A+B x}{x^3 \sqrt {b x+c x^2}} \, dx\)

Optimal. Leaf size=90 \[ \frac {4 c \sqrt {b x+c x^2} (5 b B-4 A c)}{15 b^3 x}-\frac {2 \sqrt {b x+c x^2} (5 b B-4 A c)}{15 b^2 x^2}-\frac {2 A \sqrt {b x+c x^2}}{5 b x^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {792, 658, 650} \begin {gather*} \frac {4 c \sqrt {b x+c x^2} (5 b B-4 A c)}{15 b^3 x}-\frac {2 \sqrt {b x+c x^2} (5 b B-4 A c)}{15 b^2 x^2}-\frac {2 A \sqrt {b x+c x^2}}{5 b x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^3*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*A*Sqrt[b*x + c*x^2])/(5*b*x^3) - (2*(5*b*B - 4*A*c)*Sqrt[b*x + c*x^2])/(15*b^2*x^2) + (4*c*(5*b*B - 4*A*c)
*Sqrt[b*x + c*x^2])/(15*b^3*x)

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 792

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[((d*g - e*f)*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/((2*c*d - b*e)*(m + p + 1)), x] + Dist[(m*(g*(c*d - b*e)
+ c*e*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p,
x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((L
tQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {A+B x}{x^3 \sqrt {b x+c x^2}} \, dx &=-\frac {2 A \sqrt {b x+c x^2}}{5 b x^3}+\frac {\left (2 \left (-3 (-b B+A c)+\frac {1}{2} (-b B+2 A c)\right )\right ) \int \frac {1}{x^2 \sqrt {b x+c x^2}} \, dx}{5 b}\\ &=-\frac {2 A \sqrt {b x+c x^2}}{5 b x^3}-\frac {2 (5 b B-4 A c) \sqrt {b x+c x^2}}{15 b^2 x^2}-\frac {(2 c (5 b B-4 A c)) \int \frac {1}{x \sqrt {b x+c x^2}} \, dx}{15 b^2}\\ &=-\frac {2 A \sqrt {b x+c x^2}}{5 b x^3}-\frac {2 (5 b B-4 A c) \sqrt {b x+c x^2}}{15 b^2 x^2}+\frac {4 c (5 b B-4 A c) \sqrt {b x+c x^2}}{15 b^3 x}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 54, normalized size = 0.60 \begin {gather*} -\frac {2 \sqrt {x (b+c x)} \left (A \left (3 b^2-4 b c x+8 c^2 x^2\right )+5 b B x (b-2 c x)\right )}{15 b^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^3*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*Sqrt[x*(b + c*x)]*(5*b*B*x*(b - 2*c*x) + A*(3*b^2 - 4*b*c*x + 8*c^2*x^2)))/(15*b^3*x^3)

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IntegrateAlgebraic [A]  time = 0.32, size = 60, normalized size = 0.67 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (3 A b^2-4 A b c x+8 A c^2 x^2+5 b^2 B x-10 b B c x^2\right )}{15 b^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^3*Sqrt[b*x + c*x^2]),x]

[Out]

(-2*Sqrt[b*x + c*x^2]*(3*A*b^2 + 5*b^2*B*x - 4*A*b*c*x - 10*b*B*c*x^2 + 8*A*c^2*x^2))/(15*b^3*x^3)

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fricas [A]  time = 0.41, size = 57, normalized size = 0.63 \begin {gather*} -\frac {2 \, {\left (3 \, A b^{2} - 2 \, {\left (5 \, B b c - 4 \, A c^{2}\right )} x^{2} + {\left (5 \, B b^{2} - 4 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x}}{15 \, b^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(c*x^2+b*x)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(3*A*b^2 - 2*(5*B*b*c - 4*A*c^2)*x^2 + (5*B*b^2 - 4*A*b*c)*x)*sqrt(c*x^2 + b*x)/(b^3*x^3)

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giac [A]  time = 0.22, size = 133, normalized size = 1.48 \begin {gather*} \frac {2 \, {\left (15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{3} B \sqrt {c} + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} B b + 20 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{2} A c + 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} A b \sqrt {c} + 3 \, A b^{2}\right )}}{15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )}^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(c*x^2+b*x)^(1/2),x, algorithm="giac")

[Out]

2/15*(15*(sqrt(c)*x - sqrt(c*x^2 + b*x))^3*B*sqrt(c) + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*B*b + 20*(sqrt(c)*x
- sqrt(c*x^2 + b*x))^2*A*c + 15*(sqrt(c)*x - sqrt(c*x^2 + b*x))*A*b*sqrt(c) + 3*A*b^2)/(sqrt(c)*x - sqrt(c*x^
2 + b*x))^5

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maple [A]  time = 0.05, size = 62, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (8 A \,c^{2} x^{2}-10 B b c \,x^{2}-4 A b c x +5 B \,b^{2} x +3 A \,b^{2}\right )}{15 \sqrt {c \,x^{2}+b x}\, b^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^3/(c*x^2+b*x)^(1/2),x)

[Out]

-2/15*(c*x+b)*(8*A*c^2*x^2-10*B*b*c*x^2-4*A*b*c*x+5*B*b^2*x+3*A*b^2)/x^2/b^3/(c*x^2+b*x)^(1/2)

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maxima [A]  time = 0.92, size = 106, normalized size = 1.18 \begin {gather*} \frac {4 \, \sqrt {c x^{2} + b x} B c}{3 \, b^{2} x} - \frac {16 \, \sqrt {c x^{2} + b x} A c^{2}}{15 \, b^{3} x} - \frac {2 \, \sqrt {c x^{2} + b x} B}{3 \, b x^{2}} + \frac {8 \, \sqrt {c x^{2} + b x} A c}{15 \, b^{2} x^{2}} - \frac {2 \, \sqrt {c x^{2} + b x} A}{5 \, b x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

4/3*sqrt(c*x^2 + b*x)*B*c/(b^2*x) - 16/15*sqrt(c*x^2 + b*x)*A*c^2/(b^3*x) - 2/3*sqrt(c*x^2 + b*x)*B/(b*x^2) +
8/15*sqrt(c*x^2 + b*x)*A*c/(b^2*x^2) - 2/5*sqrt(c*x^2 + b*x)*A/(b*x^3)

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mupad [B]  time = 1.09, size = 56, normalized size = 0.62 \begin {gather*} -\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (5\,B\,b^2\,x+3\,A\,b^2-10\,B\,b\,c\,x^2-4\,A\,b\,c\,x+8\,A\,c^2\,x^2\right )}{15\,b^3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^3*(b*x + c*x^2)^(1/2)),x)

[Out]

-(2*(b*x + c*x^2)^(1/2)*(3*A*b^2 + 8*A*c^2*x^2 + 5*B*b^2*x - 10*B*b*c*x^2 - 4*A*b*c*x))/(15*b^3*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{3} \sqrt {x \left (b + c x\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**3/(c*x**2+b*x)**(1/2),x)

[Out]

Integral((A + B*x)/(x**3*sqrt(x*(b + c*x))), x)

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